Question

* OWL 2 Online teaching and lea + lentActivity.do?locator assignment-take&takeAssignmentSessionLocator assignment-take [Review Topics References Use the References to access important values if needed for this question. For the following reaction, 9.39 grams of nitrogen monoxide are allowed to react with 8.26 grams of oxygen gas. nitrogen monoxide()+ oxygen(g) nitrogen dioxide(g) What is the maximum mass of nitrogen dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 6 more group attempts remaining INS HONE DEL BACK SPACE

limiting reagents ll
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Answer 1

Dear student,

NO2 (9) Answer: No1g) + 1 O₂ 18) - 2 01258 Zinitial 8.313 0.313. o 8, 258-1 x 0.313 final mass of No = 9.39 g No = ginen mass in molay mass in a glonol. - 9.39 mol - 0.313 mol. mass 2.02 = 8.26 g no2 = 8.26 mol -32 = 0.258 mol for limiting reagent dinede the nong moles of compound with their respectine s trochiometrie Coofficient. for No - 0.313 = 0.313 for O2 = 0.258 =0.516 72

o value a moles is less for No stiochiometry hence it act as limiting reagent, and reaction proceed as por No. nNo a) produce = 0.313 not mole a Noz mass q No2 = MNO2 x molar mass = 0.313 x 46 g. - 14.398 g. No 16 formular for limiting seagent es e moles of O2 remains no2 = 0.258 - ₂ x 0.313 = 0.258 -0.157 = 0.101 mol mass g02 remains a no₂ x molar mass - 0.101 * 32 g & 3.232 g.

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